SQL查询时间戳记日期字段并将所有结果按日期分组

Question

I'm trying to create a list and i'm being stumped. So I have a table that with a start_date field that holds Unix Timestamps for the date. I'm trying to query the table and extract the date from the timestamp, then any result with that date will be under that day and so on. Almost like a calendar with events on it in list view.

I'm doing this with PostgreSQL and PHP. Not sure how to write that query. I see functions like extract, date_part etc for PostgreSQL however, still unsure how to get the correct query(s).

Any help or examples would be appreciated.

Answer 1

date('d M Y', $timestamp); should give you the date, month and year of that particular UNIX timestamp.

If you would like to have it in a list view, run a select query on the database order by start_date field (ascending or descending), and fetch it in $row. Then inside the loop where you process the record one by one, do this:

$prevDate = ' ';
$currentDate = ' ';

for each ( $row ) //however it goes
{
    $currentDate = date('d M Y',$row['start_date']);
    if($prevDate != $currentDate)
    {
        echo "Day : ".$currentDate;
    }
    echo $row['other-fields'];
    $prevDate = $currentDate;
 }

However, PostgreSQL supports date datatype, and using that should simplify things at your end.

http://www.postgresql.org/docs/8.2/static/datatype-datetime.html

Answer 2

Simplest / fastest way to extract a date from a Unix epoch:

SELECT to_timestamp(my_epoch_ts)::date;

Example:

SELECT to_timestamp(1349118398)::date

Result:

2012-10-01

That's making use of the to_timestamp() function, followed by a cast to date: ::date .

More details and links at the end of this recent related answer .